Equation of all lines passing through the intersection of lines #2x + 3y -5=0# and #x+y-2=0# is given by #(2x + 3y-5) + lambda(x+y-2) = 0#, of all these lines one line is farthest from point #(7,4)# the value of #lambda# is?

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Answer 1 · 2018-02-12T12:55:40

#lambda=-4#

Let us evaluate the coordinates of the point of intersection of the lines

#2x+3y-5=0#
and

#x+y-2=0#
The equations in s and y are:

#2x+3y=5-------1#
and

#x+y=2--------2#

Multiplying--2 by 2

#2x+2y=4#

Subtracting from ----1
#=y=-1#
or
#y=1#
--2 becomes
#x+1=2#
#x=2-1#
#x=1#

All the lines pass through the point
#(x,y)=(1,1)#

The point #(7,4)# is joined with this intersection point to get the normal.

Slope of this normal is
#m1=(4-1)/(7-1)#
#m1=3/6=1/2#

The slope of the line passing through #(1,1)# needs to be perpendicular to the normal for having the farthest distance from #(7,4)#

Slope of the line farthest from #(7,4) # is
#m2=-1/(m1)=-1/(1/2)=-2#

The farthest line passes through the point #(1,1)#

Slope of the farthest line is #m=-2#

Equation of the line is
#(2x+3y-5)+lambda(x+y-2)=0#

Simplifying
#(2+lambda)x+(3+lambda)y+(-5-2lambda)=0#
#(lambda+2)x+(lambda+3)y-(2lambda+5)=0#

Slope of the line
#ax+by+c=0# is given by
#m=-a/b#

Here,
#a=lambda+2#
#b=lambda+3#

Slope, #m=-(lambda+2)/(lambda+3)#

Substituting for the slope of the line is
#m=-2#

We have
#-(lambda+2)/(lambda+3)=-2#
ie
#(lambda+2)/(lambda+3)=2#
#lambda+2=2(lambda+3)#
#lambda+2=2lambda+6#

#2lambda-lambda=2-6#
#lambda=-4#